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It is known that $1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16} + \cdots = \frac{\pi^2}{6}$ . Find the sum $1 + \frac{1}{9}+\frac{1}{25} + \frac{1}{49} + \cdots$.

What method can we use to answer this? I tried expressing the 2nd equation into 2 fractions which contain the first summation but i couldnt find one

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The usual trick: separate even and odd indices of the sum (things converge absolutely, so you can).

We have $$\frac{\pi^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \frac{1}{(2n)^2} +\sum_{n=0}^\infty \frac{1}{(2n+1)^2} $$ but $$ \sum_{n=1}^\infty \frac{1}{(2n)^2} = \sum_{n=1}^\infty \frac{1}{4n^2} = \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{4}\frac{\pi^2}{6} $$ so $$ \sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} = \frac{\pi^2}{8}\,. $$

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$S = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} +... $

Take 1/4 common from the terms whose denominator is even.

$S = 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{4} ( 1 + \frac{1}{4} + \frac{1}{9} + ... ) $

$S = 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{4} S$

$ 1 + \frac{1}{9} + \frac{1}{25} + ...= \frac{3}{4} S = (3/4)(\pi^2 /6) $

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  • $\begingroup$ Clement C. answered while I was writing this answer, but still I will leave this answer here. $\endgroup$ – Swapnil Rustagi Oct 26 '17 at 14:14

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